Classification and properties of the $\pi$-submaximal subgroups in minimal nonsolvable groups

Let $\pi$ be a set of primes. According to H. Wielandt, a subgroup $H$ of a finite group $X$ is called a $\pi$-submaximal subgroup if there is a monomorphism $\phi:X\rightarrow Y$ into a finite group $Y$ such that $X^\phi$ is subnormal in $Y$ and $H^\phi=K\cap X^\phi$ for a $\pi$-maximal subgroup $K$ of $Y$. In his talk at the well-known conference on finite groups in Santa-Cruz (USA) in 1979, Wielandt posed a series of open questions and among them the following problem: to describe the $\pi$-submaximal subgroup of the minimal nonsolvable groups and to study properties of such subgroups: the pronormality, the intravariancy, the conjugacy in the automorphism group etc. In the article, for every set $\pi$ of primes, we obtain a description of the $\pi$-submaximal subgroup in minimal nonsolvable groups and investigate their properties, so we give a solution of Wielandt's problem.


Introduction
In the paper, all groups are finite and G always denotes a finite group. Moreover, π denotes some given set of primes and π ′ is the set of primes p such that p / ∈ π. For a natural number n, let π(n) be the set of prime divisors of n and let n π be the π-share of n, that is, the greatest divisor m of n such that π(m) ⊆ π. Clearly, n = n π n π ′ and (n π , n π ′ ) = 1. For a group G, let π(G) = π(|G|). A group G is called a π-group if π(G) ⊆ π.
A subgroup H of G is said to be π-maximal if H is maximal with respect to inclusion in the set of π-subgroups of G. We denote by m π (G) the set of π-maximal subgroups of G.
A group G is called a D π -group (or write G ∈ D π ) if any two subgroups in m π (G) are conjugate. It is well-known that G ∈ D π if and only if the complete analog of the Sylow theorem holds for π-subgroups of G, that is, (1) G possesses a π-Hall subgroup (that is, a subgroup of order |G| π ), (2) every two π-Hall subgroups are conjugate in G, (3) every π-subgroup of G is contained in some π-Hall subgroup.
In particular, in a D π -group, the π-maximal subgroups are exactly the π-Hall subgroups. The Sylow theorem states that G ∈ D p for any group G and any prime p. The well-known Hall-Chunikhin theorem [3,12,13] states that a group G is solvable if and only if G ∈ D π for every set π of primes. But in general, for a given set π, the class of D π -groups is wider than the class of solvable groups since every π-group is a D π -group.
We denote by Hall π (G) the set of π-Hall subgroups of a group G. The D π -groups and the groups containing π-Hall subgroups are well investigated (see a survey [30] and next works [7,9,23,33]). In particular, it is proved that G ∈ D π if and only if every composition factor of G is a D π -groups (see [21,Theorem 7.7], [30,Theorem 6.6], [6, Chapter 2, Theorem 6.15]) and the simple D π -groups are described in [19,Theorem 3] (see also [30,Theorem 6.9]). These results are based on the description of Hall subgroups in the finite simple groups and the following nice properties of π-Hall subgroups: if N is normal and H is π-Hall subgroups in G, then HN/N ∈ Hall π (G/N) and H ∩ N ∈ Hall π (N).
The Hall-Chunikhin theorem shows that, if G is nonsolvable, then G / ∈ D π for some π. It is clear that m π (G) = ∅ and Hall π (G) ⊆ m π (G) for any group G but, in general, m π (G) Hall π (G) and it may happen that Hall π (G) = ∅. It is natural to try to find a description for π-maximal subgroups similar to ones for π-Hall subgroups. A hardness is that the π-maximal subgroups have no properties similar to the mentioned above properties of π-Hall subgroups.
For example [37, (4.2)], if X and Y are groups and X / ∈ D π for some π, then, for any π-subgroup (not only for π-maximal) K of Y , there is a π-maximal subgroup H of the regular wreath product X ≀ Y such that the image of H under the natural epimorphism X ≀ Y → Y coincides with K.
In general, the intersection of a π-maximal subgroup H with a normal subgroup N of G is not a π-maximal subgroup of N. For example, it is easy to show that a Sylow 2-subgroup H of G = P GL 2 (7) is {2, 3}-maximal in G but H ∩ N / ∈ m {2,3} (N) for N = P SL 2 (7).
But the situation with intersections of π-maximal and normal subgroups of a finite group is not so dramatic in comparing with the situation with images under homomorphisms since the following statement holds.
Proposition 1 (Wielandt-Hartley Theorem) Let G be a finite group, let N be a subnormal subgroup of G and H ∈ m π (G). Then H ∩ N = 1 if and only if N is a π ′ -group.
For the case where N normal in G, Wielandt's proof of this statement can be found in [38, 13.2], and Hartly's one in [10,Lemmas 2 and 3]. The proof of the above proposition in the general case see [24,Theorem 7].
In view of Proposition 1, it is natural to consider the following concept. According to H. Wielandt, a subgroup H of a group X is called a π-submaximal subgroup if there is a monomorphism φ : X → Y into a group Y such that X φ is subnormal in Y and H φ = K ∩ X φ for a π-maximal subgroup K of Y . We denote by sm π (X) the set of π-maximal subgroups of X.
Evidently, m π (G) ⊆ sm π (G) for any group G. The inverse inclusion does not hold in general as one can see in the above example: any Sylow 2-subgroup of P SL 2 (7) is {2, 3}-submaximal but is not {2, 3}-maximal.
In view of the Wielandt-Hartley theorem, 1 ∈ sm π (G) if and only if G is a π ′ -group 1 . Moreover, in contrast with π-maximal subgroups, the following property of π-submaximal subgroups holds: If H ∈ sm π (G) and N is a (sub)normal subgroup of G, then H ∩ N ∈ sm π (N). ( * ) This property shows that, in some sense, the behavior of π-submaximal subgroups is similar to ones of π-Hall subgroups under taking of intersections with (sub)normal subgroups. Note that properties of π-submaximal subgroups are investigated in [8,36,38] (see also [11,37]).
By using the closeness of the class of D π -group under taking extensions [21, theorem 7.7], (see also [30, theorem 6.6] and [6, Chapter 2, Theorem 6.15]) one can show that G ∈ D π if and only if every two π-submaximal subgroup of G are conjugate. In particular, if G ∈ D π , then sm π (G) = m π (G) = Hall π (G).
In view of the Hall-Chunikhin theorem, it is natural to consider some "critical" situation where G is non-solvable (and G / ∈ D π for some π) but all subgroups of G are solvable (and so they are D π -groups). In this situation, G possesses more than one conjugacy class of π-submaximal subgroups. In the paper, we consider the following problem which was possed by H. Wielandt in his talk 2 at the well-known conference on finite groups in Santa-Cruz in 1979 [37, Frage (g)]: Problem 1 (H. Wielandt, 1979) To describe the π-submaximal subgroups of the minimal nonsolvable groups. To study properties of such subgroups: conjugacy classes, the pronormality, the intravariancy 3 , the conjugacy in the automorphism group etc.
A group G is minimal nonsolvable if G is nonsolvable but every proper subgroup of G is solvable. It is well-known that G is a minimal nonsolvable group if and only if G/Φ(G) is a finite minimal simple group (that is, a non-abelian finite simple group S such that S is minimal nonsolvable), where Φ(G) is the Ftattini subgroup of G that is the intersection of the maximal subgroups of G. In 1968, J. Thompson [27,Corollary 1] proved that S is a minimal simple group if and only if S is isomorphic to a group in the following list T (we will call it as the Thompson list): (1) L 2 (2 p ) where p is a prime; (2) L 2 (3 p ) where p is an odd prime; (3) L 2 (p) where p is a prime such that p > 3 and p 2 + 1 ≡ 0 (mod 5); (4) Sz(2 p ) where p is an odd prime; Actually, the properties emphasized by Wielandt in Problem 1 play an important role in the study of π-maximal and, in particular, π-Hall subgroups. For a π-Hall subgroup H of G, the intravariancy means that H can be lifted to a π-Hall subgroup K of Aut(G) such that K ∩ Inn(G) coincides with the image of H in Inn(G) under the natural epimorphism G → Inn(G) [30,Proposition 4.8]. One of important results concerning with π-Hall subgroups of the finite simple groups is the statement that if a finite simple group S contains a π-Hall subgroup, then the number of the conjugacy classes of π-Hall subgroups in S is a bounded π-number [22, Theorem 1.1], [30,Theorem 3.4]. This statement has many consequences for arbitrary finite groups: criteria of the existence [29] and the conjugacy [28], closeness of the class of D π -groups under taking extensions, some analog of the Frattini argument for Hall subgroups [23] etc. The strongest results on Hall subgroups turned out to be formulated in term of the pronormality [7,[31][32][33]: the pronormality of Hall subgroups in the finite simple groups, the existence of a pronormal π-subgroup in any group containing π-Hall subgroups etc.
In the paper, for every set π of primes and for every minimal nonsolvable group G, we solve Problem 1. More precisely, in the first, we reduce this problem to the case where G is a minimal simple group by proving the following statement: Proposition 2 Let π be a set of primes. Let G be a finite group and let N = F (G) be the Fitting subgroup of G (that is, the greatest normal nilpotent subgroup of G). Suppose H ∈ sm π (G). Then the following statements hold: (1) m π (G/N) = {KN/N | K ∈ m π (G)}.
(4) HN/N is pronormal in G/N if and only if H is pronormal in G.

(5) H is intravariant in G if and only if the conjugacy class of
If G is a minimal nonsolvable group, then F (G) = Φ(G). Thus, for every π-submaximal subgroups H of a minimal nonsolvable group G, the image of H in the corresponding minimal simple groupG = G/Φ(G) is a π-submaximal subgroup ofG. In order to solve Problem 1, we only need to describe the π-submaximal subgroup of the minimal simple groups, that is, in the groups of the Thompson list. It is well-known [5, Corollary 1.7.10] that π(G) = π(G), and it is clear that, if π ∩ π(G) = ∅, then sm π (G) = m π (G) = {1}; if π ∩ π(G) = {p}, then sm π (G) = m π (G) = Syl p (G); and if π(G) ⊆ π, then sm π (G) = m π (G) = {G}. A description of the π-submaximal subgroup in the minimal simple groups for the remaining cases is given in the following theorem.
Theorem 1 Let π be a set of primes and S a minimal simple group. Suppose that |π ∩ π(S)| > 1 and π(S) π. Then representatives of the conjugacy classes of π-submaximal subgroups of S, the information of their structure, π-maximality, pronormality, intravariancy, and the action of Aut(S) on the set of conjugacy classes of π-submaximal subgroups can be specify in the corresponding table 1-11 below. Corollary 1.1 For every set π of primes, the π-submaximal subgroups of any minimal nonsolvable group are pronormal.
Note that our results depend on Thompson's classification of the minimal simple groups and do not depend on the classification of finite simple groups. Tables 1-11 According to [1,2,17], we use the following notation. ε denotes either +1 or −1 and the sing of this number.

Notation in
C n denotes the cyclic group of order n. E q denotes the elementary abelian group of order q where q is a power of a prime. D 2n denotes the dihedral group of order 2n, i. e. D 2n = x, y | x n = y 2 = 1, x y = x −1 . Note that D 4 ∼ = E 4 . SD 2 n denotes the semi-dihedral group of order 2 n , i. e.
S n means a symmetric group of degree n.
A n denotes the alternating group of degree n. GL n (q) denotes the general linear group of degree n over the field of order q. P GL n (q) denotes the projective general linear group of degree n over the field of order q. L n (q) = P SL n (q) denotes the projective special linear group of degree n over the field of order q.
r 1+2n + denotes the extra special group of order r 1+2n and of exponent r where r is an odd prime.
A : B means a split extension of a group A by a group B (A is normal).
A n denotes the direct product of n copies of A.
A m+n means A m : A n .
The conditions in the column "Cond." are necessary and sufficient for the existence and the π-submaximality of corresponding H. If a cell in this column is empty, then it means that the corresponding π-submaximal subgroup always exists.
In the column "H" the structure of corresponding H is given.
The conditions in the column "is not π-max. if" are necessary and sufficient for corresponding H to be not π-maximal in S. If either this column is skipped or a cell in this column is empty, then the corresponding subgroup is π-maximal.
A number n in the column "NCC" is equal to the number of conjugacy classes of πsubmaximal subgroups of S isomorphic to corresponding subgroup H and, if n > 1, then in the same column the action of Aut(S) on these classes is described.
The symbol " " in the column "Pro." means that the corresponding subgroup H is pronormal in S.
The symbol " " in the column "Intra." means that the corresponding subgroup H is intravariant in S. If a cell in this column is empty, then H is not intravariant.
In any cases H is π-maximal. The π-submaximal subgroups of S = L 2 (q), where q = 2 p , p is a prime. Case: 2 ∈ π In any cases H is π-maximal.
In any cases H is π-maximal.
In any cases H is π-maximal.
by δ * H ∼ = D 8 and H is contained in a π-maximal subgroup isomorphic to S 4 ; * * H ∼ = D 6 and H is contained in a π-maximal subgroup isomorphic to S 4 .
In any cases H is π-maximal.
Recall, a group G is said to be π-separable if G has a (sub)normal series such that every section G i−1 /G i is either a π-group or a π ′ -group.
it is sufficient to prove sm π (G) ⊆ Hall π (G). Assume that H ∈ sm π (G). Without loss of generality, we may assume that G✂✂X and The following lemma was stated without a proof in [37, 5.3]. Here we give a proof for it.
Lemma 2.8 Let π be a set of primes. Then, for a subgroup H of a non-abelian simple group S, the following conditions are equivalent.
(2) There exists a group G such that S is the socle of G, G/S is a π-group, and H = S ∩ K for some K ∈ m π (G). Proof.
(1) ⇒ (3). Take a group X of the smallest order among all groups G such that S can be embedded into G as a subnormal subgroup and H = S ∩ K for some K ∈ m π (G). Let A = S X . Then the simplicity of S implies that A is a miniml normal subgroup in X. Since X has the smallest possible order, we obtain X = KA. Now we show that K normalizes S, and as a consequence, In order to prove that N = U, we only need to show that U ∩ S = N ∩ S. Let U 0 = U ∩ S and let g 1 , . . . , g m be a right transversal of N in K. Then the subgroups S i = S g i , i = 1, . . . , m, are pairwise different and Then K normalizes V . Indeed, let x ∈ K. Since K acts by the right multiplication on {Ng 1 , . . . , Ng m }, there are a permutation σ of {1, . . . , m} and some elements t 1 , . . . , t m ∈ N such that be the natural epimorphism. Note that its restriction to S is an embedding of S into the almost simple group X with the socle S ∼ = S. It is easy to see that K ∈ m π (X) and H = K ∩ S. By the choice of X, we obtain C K (S) = 1, so X is almost simple. Thus, we can consider that X ≤ Aut(S). Let M ∈ m π (Aut(S)) such that K ≤ M. Then M ∩ S ≤ S ≤ X and K normalizes M ∩ S. This implies that M ∩ S ≤ K and M ∩ S = K ∩ S = H. ✷ Lemma 2.9 Let S be a finite simple group and let H ∈ sm π (S). Then Proof. We identify S with Inn(S). By Lemma 2.8, where q is a power of some prime p. Suppose that π is a set of primes such that 2 / ∈ π while p ∈ π. Then the following statements are equivalent.

Proof of Proposition 2
In this section, N = F (G) is the Fitting subgroup of G and H ∈ sm π (G).
Proof. Take K ∈ m π (G). Suppose, KN/N ≤ L/N where L/N is a π-group. Then L ∈ D π by Lemma 2.6. Since K ∈ m π (L), we have K ∈ Hall π (L) and KN/N ∈ Hall π (L/N) = {L/N} by Lemma 2.4. Hence, KN/N ∈ m π (G/N) and Conversely, assume that L/N ∈ m π (G/N). By Lemma 2.6, L ∈ D π and L/N = KN/N for a fixed K ∈ Hall π (L). We show that K ∈ m π (G). In fact, assume that K ≤ V for a π-subgroup V of G. Then L ≤ V N and L = V N since L/N ∈ m π (G/N). In view of K ∈ Hall π (L) and L ∈ D π , we have K = V and K ∈ m π (G). ✷ Proof. Since N is nilpotent, N has a unique π-Hall subgroup. Hence we only need to show that H ∩ N ∈ Hall π (N). By property ( * ) of π-submaximal subgroups, H ∩ N ∈ sm π (N) and sm π (N) = Hall π (N) by Lemma 2.6. ✷ (3) HN/N ∈ sm π (G/N).
Proof. Denote by φ the natural epimorphism G → G/N. We need to show that H φ ∈ sm π (G φ ). One can consider that there exists a group X such that G ✂✂ X and H = G ∩ K for some K ∈ m π (X). Denote by Y the normal closure N X of N in X.
Since N ✂ G ✂✂ X, the Fitting theorem [4, Chapter A, theorem 8 Hence G∩Y = N. Consider the restriction τ : G → X/Y to G of the natural epimorphism X → X/Y . Then the kernel of τ coincides with N. By the homomorphism theorem, there exists an injective homomorphism ψ : G φ = G/N → X/Y such that the following diagram is commutative:

Moreover, we have
where KY /Y ∈ m π (X/Y ) in view of (1). Thus H φ ∈ sm π (G φ ) by the definition of a π-submaximal subgroup. ✷ In order to prove next statements (4) and (5), we need the following lemma.
Lemma 3.1 In the above notation, HN is π-separable and H ∈ Hall π (HN).
Proof. The group HN has the subnormal series such that HN/N is a π-group, N/(H ∩ N) is a π ′ -group by (2) and H ∩ N is a π-group. Thus HN is π-separable. Moreover, H ∈ Hall π (HN) in view of Proof. If H is pronormal in G, then the pronormality of HN/N in G/N is evident. Conversely, assume that HN/N is pronormal in G/N. Let g ∈ G. We need to prove that H and H g are conjugate in H, H g .
Firstly, consider the case when g ∈ N G (HN). Then H g ≤ HN and H g ∈ Hall π (HN) by Lemma 3.1. Clearly, the subgroup H, H g of the π-separable group HN is also π-separable. Lemma 2.6 implies that the π-Hall subgroups H and H g of H, H g are conjugate in H, H g . Now consider the general case for g ∈ G.
Since HN/N is pronormal in G/N, there is y ∈ H, H g such that (HN) y = (HN) g . Hence gy −1 ∈ N G (HN) and, in view of above, there exist some z ∈ H, H gy −1 ≤ H, H g such that H z = H gy −1 . Hence H and H g are conjugate by x = zy ∈ H, H g . ✷ Proof. Recall that Aut(G) is the image in Aut(G/N) of Aut(G) under the map φ →φ whereφ : Ng → Ng φ for φ ∈ Aut(G).
If H is intravariant in G, then for every φ ∈ Aut(G), there is g ∈ G such that H φ = H g . Hence and so the conjugacy class of HN/N in G/N is invariant under Aut(G).
Conversely, assume that the conjugacy class of HN/N in G/N is invariant under Aut(G). We need to show that for every φ ∈ Aut(G) there exists g ∈ G such that Hence, H φ = H xy for some y ∈ H φ N by Lemma 2.6. ✷ 4 Proof of Theorem 1 and Corollary 1.1 We divide our proof of Theorem 1 onto three parts. In the first part (Proposition 4.1 in Section 4.1), we prove that if H is a π-submaximal subgroup of a minimal simple group S, then H can be found in that of Tables 1-11 which corresponds to given S and π. In the second part (Proposition 4.2 in Section 4.2), we prove that every H in Tables 1-11 are π-submaximal for corresponding S and prove that the information on the π-maximality, conjugacy, intravariancy and the action of Aut(S) on the conjugacy classes for this subgroups is true. Finally, in the third part (Proposition 4.3 in Section 4.3) we prove the pronormality of the π-submaximal subgroups in the minimal simple groups and, as a consequence, in the minimal nonsolvable groups (Corollary 1.1).

The classification of π-submaximal subgroups in minimal simple groups
In this section, S is a group in the Thompson list T , that is, S is one of the following groups: (1) L 2 (q) where q = 2 p , p is a prime; (2) L 2 (q) where where q = 3 p , p is an odd prime; (3) L 2 (q) where q is a prime such that q > 3 and q 2 + 1 ≡ 0 (mod 5); (4) Sz(q) where q = 2 p , p is an odd prime; We identify S with Inn(S) ∼ = S. Let π be a set of primes such that |π ∩ π(S)| > 1 and π(S) π. The following statement gives a classification of π-submaximal subgroups in the minimal simple groups. Tables 1-11 which corresponds to S and π.

Proposition 4.1 If H ∈ sm π (S) where S ∈ T , then H appears in the corresponding column in that of
Proof. Let H ∈ sm π (S). Lemma 2.8 implies that H = K ∩S for some K ∈ m π (G) where S ≤ G ≤ Aut(S) and G = KS. By Lemma 2.9, N S (H)/H is a π ′ -group and, in particular, H = 1. Since S is not a π-group, we have K < G and K ≤ M for some maximal subgroup Firstly, we consider Cases (II)-(V) keeping in mind that 1 < H ∈ Hall π (M ∩ S).
If 3 / ∈ π, then |H| = 2 and 2 ∈ π. In this case, H is contained in a Sylow 2-subgroup P of S. But 1 < N P (H)/H ≤ N S (H)/H, so N S (H)/H is not a π ′ -group, which contradicts Lemma 2.9. Thus, this case is impossible. If 2, 3 ∈ π, then H = M ∩ S and H appears in the 2-nd row of Table 6 for π + = {2}.
Assume that Case (III) holds. Then S = L 2 (q), where q = 7 and M ∩ S ∼ = D 8 . In this case, H ∈ Hall π (M ∩ S) implies that 2 ∈ π, H = M ∩ S is a Sylow 2-subgroup of S, and H appears in the 3-rd row of Table 6 for q ≡ −1 (mod 4).
Assume that Case (V) holds. Then S = L 3 (3) and M ∩ S ∼ = 3 1+2 and H coincides with a Sylow 3-subgroup of S. In this case, H appears in the 2-nd row of Table 9.
If 3 / ∈ π, then |H| = 4 and 2 ∈ π. In this case, H is contained in a Sylow 2-subgroup P of S. But 1 < N P (H)/H ≤ N S (H)/H. Hence N S (H)/H is not a π ′ -group, contrary to Lemma 2.9. Thus, this case is impossible.
But |K| = p|H| is not divisible by 3, a contradiction again. Thus, if 2 / ∈ π and 3 ∈ π, then H is not a cyclic group of order (q − 1) π . In the remaining cases, H appears in Table 3.
Suppose that 2 ∈ π. Then the maximal subgroups containing a Sylow 2-subgroup of S are D q+1 and A 4 . Hence H is a 2-group if and only if π − = {2} and H = D (q+1) 2 = D (q+1)π . In particular, such H can not be contained in E q : C 1 2 (q−1) and D q−1 . Now it is easy to see that if 3 / ∈ π, then H coincides with a π-Hall subgroup of D q−ε , where ε ∈ {+, −}; and if 3 ∈ π, then H can not coincide with a Sylow 2-subgroup P by Lemma 2.9 in view of N S (P ) ∼ = A 4 . Hence H must coincides with one of the following groups E q : C 1 2 (q−1)π , D (q−1)π , D (q+1)π or A 4 . In all these cases, H appears in Table 4.
Suppose that 2 / ∈ π. Since | Aut(S) : S| = 2, H must be a π-maximal subgroup of S by Lemma 2.8. If H is contained in one of A 4 or S 4 , then |H| = 3 and, since any Sylow 3-subgroup of S is cyclic, H is contained in D q−ε with q ≡ ε (mod 3). Moreover, if q ∈ π, then the unique π-Hall subgroup of D q−1 is not π-maximal in S. Indeed, this subgroup is a cyclic π + -Hall subgroup of S. Lemma 2.5 implies that this subgroup is conjugate to a subgroup in some Frobenius group C q : C 1 2 (q−1) ≤ S and normalizes a Sylow q-subgroup of S. Hence, if 2 / ∈ π, then H appears in Table 5. Suppose that 2 ∈ π. In order to show that H appears in Table 6, it is sufficient to show that a π-Hall subgroup H of D q−ε is not π-submaximal in S if π ε = {2}, 3 ∈ π and q ≡ ε (mod 8). Indeed, if q ≡ −ε (mod 8), then |H| = (q − ε) 2 = 2 < |S| 2 and H < P for some Sylow 2-subgroup P of S, which contradicts Lemma 2.9; if q ≡ ±3 (mod 8), then H is an elementary abelian Sylow 2-subgroup of S, and |H| = 4, which contradicts Lemma 2.9 since, by the Sylow theorem, H is contained as a normal subgroup in a maximal subgroup of S isomorphic to A 4 and this subgroups is a π-group.
Note that S has exactly one conjugacy class of cyclic subgroups of order q − 1 (see the character table of S in [25, theorem 13]). Hence, if 2 / ∈ π, then any π-Hall subgroup of E 1+1 q : C (q−1) is contained as a π-Hall subgroup in some maximal subgroup of kind C (q−1)π : C 4 . Thus, in the case when 2 / ∈ π, H appears in Table 7. Now assume that 2 ∈ π. If one of the sets π 0 , π + , π − is empty, then any π-Hall subgroup of the respective subgroups: or C (q−1)π : C 4 , or D 2(q−r+1) , or D 2(q−r+1) , is a 2group, but is not a Sylow 2-subgroup. Then Lemma 2.9 implies that case where H is a π-Hall subgroup in one of these maximal subgroup of S is impossible. In the remaining cases, H appears in Table 8.
Suppose that π ∩ π(S) = {3, 13}. Then a π-Hall subgroup H of E 3 2 : GL 2 (3) coincides with a Sylow 3-subgroup 3 1+2 + of S and appears in Table 9. A π-Hall subgroup H of C 13 : C 3 is C 13 : C 3 itself and H appears in Table 9. A π-Hall subgroup H of S 4 is of order 3 and Lemma 2.9 implies that this case is impossible.
Suppose that π ∩ π(S) = {2, 13}. Then a π-Hall subgroup H of E 3 2 : GL 2 (3) coincides with a semi-dihedral Sylow 2-subgroup SD 16 of both E 3 2 : GL 2 (3) and S, and H appears in Table 10. A π-Hall subgroup H of C 13 : C 3 is cyclic of order 13 and appears in Table 10. A π-Hall subgroup H of S 4 is of order 8 and is not a Sylow 2-subgroup of S. We may exclude this case by Lemma 2.9.
Finally, suppose that π ∩ π(S) = {2, 3}. Then a π-Hall subgroup H of E 3 2 : GL 2 (3) is E 3 2 : GL 2 (3) itself and H appears in Table 11. A π-Hall subgroup H of C 13 : C 3 is cyclic of order 3 and Lemma 2.9 implies that this case is impossible. A π-Hall subgroup H of S 4 is S 4 itself and it appears in Table 11. ✷ 4.2 The π-submaximality, π-maximality, conjugacy, and intravariancy for subgroups in Tables 1-10 In this section, as in above, S is a group in the Thompson list T . We prove that the converse of Proposition 4.1 holds, that is, the subgroups placed in Tables 1-10 are π-submaximal under corresponding conditions. Moreover, we prove that the information about the π-maximality, the conjugacy, the action of Aut(S) on the conjugacy classes of this subgroups, and the intravariancy in this tables is correct.
Proposition 4.2 Let S ∈ T , π be a set of primes such that |π ∩ π(S)| > 1 and π(S) π, and let H be a subgroup of S placed in corresponding column of one of Tables 1-11 which corresponds to S. Then (1) H ∈ sm π (S).
In this case, H is a Sylow 2-subgroup of S, H ∼ = D 8 , and H < S 4 ⋖ S.
Consider the case where π ∩ π(q − ε) = {2} and q ≡ −ε7 (mod 48). Then q ≡ ε (mod 8). This means that H coincides with a Sylow 2-subgroup of S and, moreover, S contains a maximal subgroup isomorphic to S 4 by Lemma 2.1. Since q ≡ −7ε (mod 16), we have |H| = (q − ε) π = 8 = |S 4 | 2 . The Sylow theorem implies that H is conjugate to a Sylow 2-subgroup of S 4 . Thus, H is not π-maximal in S in view of 2, 3 ∈ π and S 4 is a π-group. Now consider the case where π ∩ π(q − ε) = {2, 3} and q ≡ ε7, ε31 (mod 72). Then q ≡ −ε (mod 8). It means that S contains a maximal subgroup M isomorphic to S 4 by Lemma 2.1. But it is easy to calculate that |H| = (q − ε) π = 6 and so H ∼ = S 3 . Moreover, H contains a Sylow 3-subgroup P of S since |S| 3 = 3. It follows that H is contained in N S (P ). By considering the maximal subgroups of S given in Lemma 2.1, it is easy to see that N S (P ) = D q−ε , so H is a {2, 3}-Hall subgroup of N S (P ). By the Sylow theorem, we can assume that P < M. Since N M (P ) ∼ = S 3 , we obtain N M (P ) ∈ Hall {2,3} (N S (P )). The Hall theorem and the solvability of N S (P ) imply that H is conjugate with N M (P ) and H is not π-maximal in S since 2, 3 ∈ π and S 4 is a π-group.
In order to complete the proof of (2), it is sufficient to show that, in the remaining cases, H is a π-Hall subgroup in every maximal subgroup M of S containing H. By Lemma 2.1, an easy calculation show that one of the following cases holds: (I) up to isomorphism, there is a unique maximal subgroup M of S such that |H| divides |M|, H ≤ M and H is a π-Hall subgroup of M.
(II) S = L 2 (q) where q = 2 p for a prime p, 2 / ∈ π, H = C (q−1)π , and, up to isomorphism, there are exactly two maximal subgroup M of S such that |H| divides |M|, namely either M = E q : C (q−1) or M = D 2(q−1) . In this case, H is a π-Hall subgroup of every maximal subgroup containing H.
(III) S = L 2 (q) where q = 2 p for a prime p, 2 ∈ π, H = D 2(q−1)π , and, up to isomorphism, there are exactly two maximal subgroup M of S such that |H| divides |M|, namely either M = E q : C (q−1) or M = D 2(q−1) . In this case H is not isomorphic to a subgroup of E q : C (q−1) and is contained only in M = D 2(q−1) . Hence H is a π-Hall subgroup of every maximal subgroup containing H.
(IV) S = L 2 (q) where either q = 3 p for odd prime or q > 3 is a prime, 2 / ∈ π, H = C (q−1)π and, up to isomorphism, there are exactly two maximal subgroup M of S such that |H| divides |M|, namely either M = E q : C 1 2 (q−1) or M = D q−1 . In this case, H is a π-Hall subgroup of every maximal subgroup containing H.
(V) S = Sz(q) where q = 2 p for a prime p, 2 / ∈ π, H = C (q−1)π , and, up to isomorphism, there are exactly two maximal subgroup M of S such that |H| divides |M|, namely either M = E 1+1 q : C (q−1) or M = C q−1 : C 4 . In this case, H is a π-Hall subgroup of every maximal subgroup containing H.
(VI) S = Sz(q) where q = 2 p for a prime p, 2 ∈ π, H = C (q−1)π : C 4 , and, up to isomorphism, there are exactly two maximal subgroup M of S such that |H| divides |M|, namely either M = E 1+1 q : C (q−1) or M = C q−1 : C 4 . In this case H is not isomorphic to a subgroup of E 1+1 q : C (q−1) and is contained only in M = C q−1 : C 4 . Hence H is a π-Hall subgroup of every maximal subgroup containing H.
(VIII) S = L 2 (q) where q is a prime such that q > 3 and q 2 + 1 ≡ 0 (mod 5), 2, 3 ∈ π, H ∈ {A 4 , S 4 }, |H| divides q − ε for some ε ∈ {+, −} and, up to isomorphism, there are exactly two maximal subgroup M of S such that |H| divides |M|, namely either M = D q−ε or M ∈ {A 4 , S 4 }. In this case H is not isomorphic to any subgroup of D q−ε , so H itself is maximal and coincide with its π-Hall subgroup. Thus, in order to complete the proof of (2), we need to consider Case (VII): S = L 2 (q) where q is a prime such that q > 3 and q 2 + 1 ≡ 0 (mod 5), 2 ∈ π, H = D (q−ε)π for some ε ∈ {+, −}, |H| divides 24, and, up to isomorphism, there are exactly two maximal subgroup M of S such that |H| divides |M|, namely either M = D q−ε or M ∈ {A 4 , S 4 }. It is clear that H is a π-Hall subgroup of D q−ε . We need to show that either H is not contained in M ∈ {A 4 , S 4 }, or H is a π-Hall subgroup in such M, or one of the exceptional cases (2a) and (2b) holds. Note that we have the condition in Table 6 for H that ether π ε = {2}, or 3 / ∈ π, or q ≡ ε (mod 8).
Thus, exceptional case (2b) holds. Thus, we can consider that π ε = {2}. Suppose that 3 / ∈ π. Since |H| divides 24, we have H = D 8 or H = D 4 = C 2 × C 2 . In the first case, M = S 4 and H is a π-Hall subgroup of both D q−ε and S 4 , and in the second case, M = A 4 and H is π-Hall in both D q−ε and A 4 .
This completes the proof of (2). Now we prove (1). In view of (2), it is sufficient to prove that if one of the exceptional cases (2a)-(2d) in (2) holds, then H = K ∩ S for some K ∈ m π (G) where G = Aut(S).
Prove (3). By Lemma 2.1 (Tables 14 and 16), if one of the exceptional cases (3a) or (3b) holds, then H is a maximal subgroup of corresponding S, any maximal subgroup of S which is not isomorphic to H does not contain a subgroup isomorphic to H, and S has exactly two conjugacy classes of (maximal) subgroups isomorphic to H interchanged by every non-inner automorphism of S.
We need to prove that in the remaining cases, every K ∈ sm π (S) isomorphic to H is conjugate to H. By (2), if there is a maximal subgroup M of S containing both H and K, then H, K ∈ Hall π (M). The solvability of M implies that H and K are conjugate in M in view the Hall theorem. Hence we can consider that H and K are contained in non-conjugate maximal subgroups M and N, respectively.
If M and N are isomorphic, then Lemma 2.1 implies that either (a) S = L 2 (q), q > 3 is a prime, and M ∼ = N ∼ = S 4 , or (b) S = L 3 (3) and M ∼ = N ∼ = E 3 2 : GL 2 (3). Since H ∈ Hall π (M) and K ∈ Hall π (N) in view of (2) and H = M and K = N (if not, one of the exceptional cases (3a) or (3b) holds), H and K are Sylow 2-or 3-subgroups of M and N, respectively. Now Lemma 2.9 and H, K ∈ sm π (S) imply that H, K are Sylow subgroups in S and they are conjugate by the Sylow theorem.
Consider the case where M and N are non-isomorphic maximal subgroups of S. In this case, |H| = |K| divides (|M|, |N|) and, by the arguments similar above, we can consider that the numbers |H| = |K| and (|M|, |N|) are not powers of primes. Consider all possibilities for S.
Let S = L 2 (q) where q = 2 p , p is a prime. Since (|M|, |N|) is not a power of prime, by the information in Table 12 we can consider that M = E q : C (q−1) and N = D 2(q−1) . If 2 ∈ π, then the π-Hall subgroups of M and N are non-isomorphic. Hence 2 / ∈ π. In this case, both H and K are abelian π-Hall subgroups of S and they are conjugate by Lemma 2.5.
Let S = L 2 (q), where q = 3 p and p is an odd prime. Note that the order of E q : C 1 2 (q−1) is odd and the orders of D q−1 and D q+1 are not divisible by 3. Now it easy to see from Table 13 that the condition that (|M|, |N|) is not a power of prime implies that we can consider that M = E q : C 1 2 (q−1) , N = D q−1 , and 2 / ∈ π. In this case, both H and K are abelian π-Hall subgroups of S and they are conjugate by Lemma 2.5.
Let S = L 2 (q) where q is a prime such that q > 3 and q 2 + 1 ≡ 0 (mod 5). The case when one of M and N is E q : C 1 2 (q−1) and other one is D q−1 can be argued similarly as the previous case, and as (|M|, |N|) is not a power of prime, we can consider that M ∈ {A 4 , S 4 }. Hence 2, 3 ∈ π. If M = A 4 , then H = M (if not, |H| is a power of 2 or 3), but the maximal subgroups of S of the other types contain no subgroups isomorphic to A 4 . Hence M = S 4 . Since the other maximal subgroups of S do not contain subgroups isomorphic to S 4 and H is not a power of a prime, we have that H ∼ = S 3 . Since a maximal subgroup N of S contains a subgroup K isomorphic to H ∼ = S 3 , it is easy to see from Table 14 that N = D q−ε for some ε ∈ {+, −} and in view of K ∈ Hall π (N) we have that |S| 3 = |N| 3 = 3. Hence N is the normalizer of a Sylow 3-subgroup P of S and K ∈ Hall π (N S (P )). But H ∼ = K means that H is also a π-Hall subgroup of some (solvable) normalizer of a Sylow 3-subgroup Q of S. Hence by the Sylow and Hall theorems, H and K are conjugate.
Let S = Sz(q), where q = 2 p , p is an odd prime. This case can be investigated without essential changes as the case S = L 2 (q), q = 2 p .
In this case, H is pronormal as a Sylow subgroup of S.